A password shall begin with 3 different capital letters, followed by 3 to 5 digits between 0 to 9.

a)How many possibilities of passwords are there?

b)How many passwords with the character string "B4" are there?

(with binomial Coefficient = $n \choose k$)

a)For the 3 first digits I have $26\choose 3$ possibilities. Then they are followed by 3 to 5 digits, and I don't know how to approach...

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a) There are 3 different cases for this. For when there are 3 numbers, 4 numbers, 5 numbers

There are 3 different letters with 26 to choose from. Hence $\displaystyle\binom{26}{3}$ but this doesn't take into account order so we multiply by $3!$. Then for case 3, 4, 5. Since there are 10 numbers to choose from. We multiply $\displaystyle\binom{26}{3}\times 3!$ by $10^3$, $10^4$, $10^5$ respectively for cases when we use 3, 4, 5 digits. I assumed here that the numbers can be repeated.

b) Since the character string is "B4" (a letter then a number) then B is in the 3rd position, and 4 is in the 4th position.

$_,_,B,4,_,_$ is our format. With permutations $25\times24\times1\times1\times10\times10$

Since B and 4 are fixed we have 1 for the 3rd and 5th position. And $25, 24$ since we alread chose $B$ for the 3rd character.

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$\binom{26}{3}\cdot 3!$ seems an awefully roundabout way of writing $26\cdot 25\cdot 24$. But then again, the question was asked with binomial coefficients in mind. –  Arthur Nov 22 '12 at 12:51
So order means, that every number or letter has its assigned place/digit? –  phil Nov 22 '12 at 12:52
@user49421 Yes. The binomial coefficients are used only when you don't care which is first, which is second and so on, you just care which are picked and which aren't. Of course, you can use binomial coefficients and then take care of ordering afterwards, as is done in the answer above. –  Arthur Nov 22 '12 at 12:53
Yes. I actually shouldv'e just written $26P3$ Lol. :) But it was just to clarify the point of order. Yes. For each permutation of letters and numbers, every letter has a designated position. For example if we want to put order, ABC is different from BAC. But if we don't consider order, ABC would be the same as BAC. But since we are taking passwords into consideration. We should need to take into account order to make guessing more "complex" Haha :) –  Keith Reyes Nov 22 '12 at 12:56
Now I have understood a) :), and because of this order of place I also have to write 10^k with k=place/digit? –  phil Nov 22 '12 at 13:02

a) You can pick the first letter in 26 ways (1 out of 26 letters). The next one will be from the remaining 25 letters, so ways of picking it would be 25. Similarly, the 3rd letter can be picked in 24 ways. So picking the first 3 letters can be done in 26x25x24 ways. For the numbers, ways of picking each digit can be done in 10 ways. As having 3 digits, 4 digits and 5 digits are mutually exclusive events, we can add the ways of generating each. Thus, for 3 digits, we have 26x25x24x$10^3$ ways. Similarly for 4 and 5 digits, giving the total number of ways to be: 26x25x24x($10^3 + 10^4 + 10^5$).

b) You need to fix the 3rd char to B and the first digit to 4 (that is the only place where a number follows a letter). Select the rest from the remaining 25 letters like above, to give the number of ways as: 25x24x($10^2 + 10^3 + 10^4$).

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For b) we already chose 1 letter. So there are 25 letters to choose from for the first position. I think. =) –  Keith Reyes Nov 22 '12 at 12:40
@KeithReyes Yep, you are right. Thanks! Fixed. –  Paresh Nov 22 '12 at 12:45

a) How about ${10}^3+{10}^4+{10}^5$? (EDIT: Sorry, I first had assumed the digits shold be different as well)

b) Hint: Where can "B4" occur at all?

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that means I only have to add from the third to the 5th digit this sum.Ok and for b) it is only one possibility, because it's the only one where a letter is followed by a digit? –  phil Nov 22 '12 at 12:40
But in a) didn't take into account the permutations for the letters. The sum should be multiplied by $26\times25\times24$ –  Keith Reyes Nov 22 '12 at 12:43